$a)$

$x^3-7x^2+12x=0$

$x\left(x^2-7x+12\right)=0$

$x=0\vee x^2-7x+12=0$

$\Delta ={\left(-7\right)}^2-4\cdot 1\cdot 12=49-48=1$

$\sqrt{\Delta }=1$

$x=\frac{7-1}{2}=3\vee x=\frac{7+1}{2}=4$

$\underline{x\in \left\{0,3,4\right\}}$